Question

The structures of \(BCl_3\) and \(BCl_3 \cdot NH_3\), respectively are

• A. Planar trigonal and Tetrahedral
• B. Planar trigonal and Pyramidal
• C. Pyramidal and Tetrahedral
• D. Pyramidal and Pyramidal

Hint:

\(BCl_3\) is electron deficient and has \(sp^2\) hybridization with trigonal planar geometry. When it accepts a lone pair from \(NH_3\), boron completes its octet and becomes \(sp^3\) hybridized with tetrahedral geometry.

Answer 1

The Correct Option is A

Step 1: Determine the structure of \(BCl_3\).
In \(BCl_3\), boron has three valence electrons and forms three \(B-Cl\) bonds. The central boron atom has: \[ 3 \text{ bond pairs} \] and \[ 0 \text{ lone pairs} \] According to VSEPR theory, the steric number is \[ 3 \] Hence, boron undergoes \[ sp^2 \] hybridization.
Therefore, the geometry of \(BCl_3\) is \[ \boxed{\text{Trigonal Planar}} \] with bond angle \[ 120^\circ \]

Step 2: Understand the formation of \(BCl_3 \cdot NH_3\).
\(BCl_3\) is electron deficient and acts as a Lewis acid. Ammonia (\(NH_3\)) contains a lone pair on nitrogen and acts as a Lewis base. Nitrogen donates its lone pair to boron: \[ BCl_3 + NH_3 \rightarrow BCl_3 \cdot NH_3 \] forming a coordinate covalent bond.

Step 3: Determine the geometry around boron in \(BCl_3 \cdot NH_3\).
After accepting the lone pair from nitrogen, boron is surrounded by: \[ 3 \text{ B-Cl bonds} \] and \[ 1 \text{ coordinate bond } (B \leftarrow NH_3) \] Thus, boron has \[ 4 \text{ bond pairs} \] and no lone pair.
The hybridization changes to \[ sp^3 \] which gives a \[ \boxed{\text{Tetrahedral}} \] geometry around boron.

Step 4: Compare with the given options.
\[ BCl_3 \rightarrow \text{Trigonal Planar} \] \[ BCl_3 \cdot NH_3 \rightarrow \text{Tetrahedral} \] Only option (1) matches these geometries.

Step 5: Final conclusion.
Hence, the structures of \[ BCl_3 \] and \[ BCl_3 \cdot NH_3 \] are respectively \[ \boxed{\text{Planar trigonal and Tetrahedral}} \] Therefore, option (1) is correct.